Problem: Equilateral triangle $ABC$ and square $BCDE$ are coplanar, as shown. What is the number of degrees in the measure of angle $CAD$?

[asy]
size(70);
draw((0,0)--(20,0)--(20,20)--(0,20)--cycle);
draw((0,20)--(10,37.3)--(20,20));
draw((10,37.3)--(20,0));
label("$A$",(10,37.3),N);
label("$B$",(0,20),W);
label("$C$",(20,20),E);
label("$D$",(20,0),E);
label("$E$",(0,0),W);
[/asy]
Answer: First, observe that $AC=CD$.  Therefore, triangle $ACD$ is isosceles and $\angle CAD$ is congruent to $\angle CDA$.  Also, $m\angle ACD=m\angle ACB+m\angle BCD=60^\circ+90^\circ=150^\circ$.  Since the three angles of triangle $ACD$ sum to 180 degrees, we have  \begin{align*}
m\angle CAD+m\angle CDA+150^\circ&=180^\circ \implies \\
2m\angle CAD&=30^\circ\implies \\
m\angle CAD&=\boxed{15} \text{ degrees}.
\end{align*}